A 2 kg block is thrown at 40 m/s at an unknown angle from the top of a the groun

Question

A 2 kg block is thrown at 40 m/s at an unknown angle from the top of a the ground at 50 m/s. Determine the height of the building. Ignore air resistance. A baseball player slides towards home plate from a 5 m distance and with initial speed m/s. If the coefficient of kinetic friction between the player and the ground is 0.48, determine the speed with which he hits the plate. A block of mass m hits an identical, stationary block after they slide together for 1.8m before stopping on a horizontal with coefficient of kinetic friction 0.25. Determine the initial speed of the moving block. A ball of mass 1 kg moves in a vertical circle while attached to a light string of length 0.3m fixed at its center. Assume that the only force acting on the ball are its weight and the tension of the string. Determine the minimum speed the ball must have at point A (see Figure 4) so as to complete the circle by reaching the top. For the same ball in problem 4, if the tension at the bottom point B is eight times the weight of the ball, determine the tension at point D (see Figure 5). Ball A of mass m and initial velocity v collides head on elastically with ball B, of mass 3m and initially stationary. Determine the final velocities of A and B in terns of v.

Explanation / Answer

1) Applying energy conservation

(1/2)mu2 + mgh = (1/2)mv2

=> h = (1/2g)(v2 - u2)

=> h = 45.87 metres

-------------------------------------------------------------------------------------------------------------------------

2) applying energy conservation

(1/2)mu2 + Wfriction = (1/2)mv2

=> (1/2)mu2 - (mg*s) = (1/2)mv2

=> (1/2)*m*(82) - (0.48*m*g)(5) = (1/2)mv2

=> v = 4/11 m/s

-------------------------------------------------------------------------------------------------------------------------

3)

Let 'u' be the velocity before collision, and 'v' be the velocity after collision

Applying momentum conservation

mu = mv + mv

=> v = u/2

Applying energy conservation

(1/2)(2m)(v2) + Wfriction = 0

=> m(u2/4) - (mg*s) = 0

=> u2/4 = 0.25*9.81*1.8

=> u = 4.2 m/s

------------------------------------------------------------------------------------------------------------------

4) and 5) Figure 4 is missing

--------------------------------------------------------------------------------------------------------------------

6) After collison, let 'v1' be the velocity made of m 'v2' be the velocity of 3m

Conserving momentum.

mu = m(v1) + 3m(v2)

=> u = v1 + 3v2

Since the collision is elastic, energy will also be conserved

(1/2)mu2 = (1/2)m(v12) + (1/2)(3m)(v22)

=> u2 = v12 + 3v22

Solve the above two equations for v1 and v2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.